UVa 10891 Game of Sum (DP)

题意：给定一个长度为n的整数序列，两个人轮流从左端或者右端拿数，A先取，问最后A的得分-B的得分的结果。

析：dp[i][j] 表示序列 i~j 时先手得分的最大值，然后两种决策，要么从左端拿，要么从右端拿，肯定是拿的是最大的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn][maxn], sum[maxn], a[maxn];
int f[maxn][maxn], g[maxn][maxn];

int main(){
    while(scanf("%d", &n) == 1 && n){
        for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
        sum[0] = 0;
        for(int i = 1; i <= n; ++i){
            f[i][i] = g[i][i] = dp[i][i]  = a[i];
            sum[i] = sum[i-1] + a[i];
        }
        for(int l = 1; l < n; ++l)
            for(int i = 1; i+l <= n; ++i){
                int j = i + l;
                int mmin = 0;
                mmin = min(mmin, f[i+1][j]);
                mmin = min(mmin, g[i][j-1]);
                dp[i][j] = sum[j] - sum[i-1] - mmin;
                f[i][j] = min(dp[i][j], f[i+1][j]);
                g[i][j] = min(dp[i][j], g[i][j-1]);
            }
        printf("%d
", 2*dp[1][n] - sum[n]);
    }
    return 0;
}